define and proof midpoint theorem
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Answered by
5
Hey there!
→ Given: AD = DB and AE = EC.
→ To Prove: DE ∥ BC and DE = 1/2 BC.
→ Construction: Extend line segment DE to F such that DE = EF.
→ Proof: In △ ADE and △ CFE
AE = EC (given)
∠AED = ∠CEF (vertically opposite angles)
DE = EF (construction)
Hence,
△ ADE ≅ △ CFE (by SAS)
Therefore,
∠ADE = ∠CFE (by c.p.c.t.)
∠DAE = ∠FCE (by c.p.c.t.)
and AD = CF (by c.p.c.t.)
The angles ∠ADE and ∠CFE are alternate interior angles assuming AB and CF are two lines intersected by transversal DF.
Similarly, ∠DAE and ∠FCE are alternate interior angles assuming AB and CF are two lines intersected by transversal AC.
Therefore, AB ∥ CF
So, BD ∥ CF
and BD = CF (since AD = BD and it is proved above that AD = CF)
Thus, BDFC is a parallelogram.
By the properties of parallelogram, we have
DF ∥ BC
and DF = BC
DE ∥ BC
and DE = 1/2 BC (DE = EF by construction)
Hence Proved... !
Cheers!
→ Given: AD = DB and AE = EC.
→ To Prove: DE ∥ BC and DE = 1/2 BC.
→ Construction: Extend line segment DE to F such that DE = EF.
→ Proof: In △ ADE and △ CFE
AE = EC (given)
∠AED = ∠CEF (vertically opposite angles)
DE = EF (construction)
Hence,
△ ADE ≅ △ CFE (by SAS)
Therefore,
∠ADE = ∠CFE (by c.p.c.t.)
∠DAE = ∠FCE (by c.p.c.t.)
and AD = CF (by c.p.c.t.)
The angles ∠ADE and ∠CFE are alternate interior angles assuming AB and CF are two lines intersected by transversal DF.
Similarly, ∠DAE and ∠FCE are alternate interior angles assuming AB and CF are two lines intersected by transversal AC.
Therefore, AB ∥ CF
So, BD ∥ CF
and BD = CF (since AD = BD and it is proved above that AD = CF)
Thus, BDFC is a parallelogram.
By the properties of parallelogram, we have
DF ∥ BC
and DF = BC
DE ∥ BC
and DE = 1/2 BC (DE = EF by construction)
Hence Proved... !
Cheers!
Yuichiro13:
:p
Answered by
2
Heya User,
Midpoint theorem states that :->
--> " The line joining the mid-points of the two sides of a triangle is always parallel to the third side and is half of it "
--> For Class 8 I'd suppose :->
--> Brother Ishan has given an awesome answer... Here's a different one if uh want to... Umm.. If uh hate constructions that is....
--> Notations used in my answer :->
---> [ ABC ] --> Area of ABC ...
Proof :->
---> Take a ΔABC, and D, E as mid-points on AB, AC respectively..
--> Clearly, we know that [ ABE ] = [ CBE ] = 1/2 [ ABC ] --> (i)
Similarly, --> [ ACD ] = [ BCD ] = 1/2 [ ABC ] ---> (ii)
Equating (i) and (ii) via. the 1/2 [ ABC ],
--> [ ABE ] = [ ACD ]
=> [ ADE ] + [ BDE ] = [ ADE ] + [ CDE ]
=> [ BDE ] = [ CDE ]
--> and we know that since they are bound b/w same lines 'DE' and BC and have same area, they must be lying between parallels and hence, DE || BC..
Now, for the second part, that DE = 1/2 BC thingy... draw F as the mid-point of BC and join EF... From the above proof [ =_= that I gave ] , we have :->
--> DE || BF and and BD || EF .. Hence, BDEF is a ||gm :->
--> DE = BF = 1/2 BC and we're done...
Ummm, for the sake of understanding, I've made this a bit longer or the proof is quite short in comparizon... xD ^_^ Enjoy..
Midpoint theorem states that :->
--> " The line joining the mid-points of the two sides of a triangle is always parallel to the third side and is half of it "
--> For Class 8 I'd suppose :->
--> Brother Ishan has given an awesome answer... Here's a different one if uh want to... Umm.. If uh hate constructions that is....
--> Notations used in my answer :->
---> [ ABC ] --> Area of ABC ...
Proof :->
---> Take a ΔABC, and D, E as mid-points on AB, AC respectively..
--> Clearly, we know that [ ABE ] = [ CBE ] = 1/2 [ ABC ] --> (i)
Similarly, --> [ ACD ] = [ BCD ] = 1/2 [ ABC ] ---> (ii)
Equating (i) and (ii) via. the 1/2 [ ABC ],
--> [ ABE ] = [ ACD ]
=> [ ADE ] + [ BDE ] = [ ADE ] + [ CDE ]
=> [ BDE ] = [ CDE ]
--> and we know that since they are bound b/w same lines 'DE' and BC and have same area, they must be lying between parallels and hence, DE || BC..
Now, for the second part, that DE = 1/2 BC thingy... draw F as the mid-point of BC and join EF... From the above proof [ =_= that I gave ] , we have :->
--> DE || BF and and BD || EF .. Hence, BDEF is a ||gm :->
--> DE = BF = 1/2 BC and we're done...
Ummm, for the sake of understanding, I've made this a bit longer or the proof is quite short in comparizon... xD ^_^ Enjoy..
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